Question

You have 2 fair coins and one coin with heads on both sides. You pick a coin at random and toss it twice. If it lands heads up on both tosses, the probability it also lands heads up on a third toss can be express in the form A/B, where A and B are relatively prime positive integers (i.e. the greatest common divisor is 1). Compute A + B.

Answer 1

There are two fain coins and one coin with heads on both side, for fair coin P[ getting head while tossing] = 1/2 For coin with bead bothside P I getting head ] = 1. For random pick of any coin P[ getting any coin] = $3. Now from the 3 coin, if we take a random pick- PI Getting head while tossing = PI Getting head I 181 fain coin comes up] PI 1st fain coin comes] +P [ Getting head and fair coin connes] P[and fair coin comes] +P [ Getting head) coin both faces head ] P[ coin with both face head] voitt = (x) + (x5) + (1x1) Now olor objective is to find - চকমতে ৮ P[ It lands head on and toss | Both toss head] = PI It lands head on 3rd toss in Both toss head] P[ Both toss head] I since P(AIB) & P[ three toss head ocurs] = P(ARB) P[ Both PB) toss head occurs]

Therefore, = PI & lands head on all the three toss] Pl lands head on both toss] 1 2/x % 2 Therefore, with GOD (AIB) = 1 A=2 B = 3 . At B=5 cans